Question

How do you express $\dfrac{1}{{{x^4} + 1}}$ in partial fractions?

Answer 1

Hint: This problem deals with reducing the given complex fraction into partial fractions. In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction is an operation that consists of expressing the fraction as a sum of a polynomial and one or several fractions with a simpler denominator.

Complete step-by-step answer:
Given a complex fraction which is as given below:
\[ \Rightarrow \dfrac{1}{{{x^4} + 1}}\]
Now consider the denominator of the above given partial fraction, as shown below:
$ \Rightarrow {x^4} + 1$
Now this a polynomial of degree four, and this can be reduced into factors of polynomials of degree two, as shown below:
$ \Rightarrow {x^4} + 1 = \left( {{x^2} - \sqrt 2 x + 1} \right)\left( {{x^2} + \sqrt 2 x + 1} \right)$
Now considering the given complex fraction as given below:
$ \Rightarrow \dfrac{1}{{{x^4} + 1}} = \dfrac{1}{{\left( {{x^2} - \sqrt 2 x + 1} \right)\left( {{x^2} + \sqrt 2 x + 1} \right)}}$
Now splitting the fraction on the right hand side of the equation, as shown:
$ \Rightarrow \dfrac{1}{{{x^4} + 1}} = \dfrac{{Ax + B}}{{\left( {{x^2} - \sqrt 2 x + 1} \right)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + \sqrt 2 x + 1} \right)}}$
Now simplifying the right hand side of the equation:
$ \Rightarrow \dfrac{1}{{{x^4} + 1}} = \dfrac{{Ax + B\left( {{x^2} + \sqrt 2 x + 1} \right) + Cx + D\left( {{x^2} - \sqrt 2 x + 1} \right)}}{{\left( {{x^2} - \sqrt 2 x + 1} \right)\left( {{x^2} + \sqrt 2 x + 1} \right)}}$
We know that denominators are equal, now equating the numerators, as shown:
\[ \Rightarrow \left( {Ax + B} \right)\left( {{x^2} + \sqrt 2 x + 1} \right) + \left( {Cx + D} \right)\left( {{x^2} - \sqrt 2 x + 1} \right) = 1\]
\[ \Rightarrow A{x^3} + \sqrt 2 A{x^2} + Ax + B{x^2} + \sqrt 2 Bx + B + C{x^3} - \sqrt 2 C{x^2} + Cx + D{x^2} - \sqrt 2 Dx + D = 1\]
Now grouping the like terms and the unlike terms together:
\[ \Rightarrow \left( {A + C} \right){x^3} + \left( {\sqrt 2 A + B - \sqrt 2 C + D} \right){x^2} + \left( {A + \sqrt 2 B + C - \sqrt 2 D} \right)x + \left( {B + D} \right) = 1\]
Here on the right hand side, there is no ${x^3}$ term or ${x^2}$ term or an $x$ term, only a constant which is 1.
Hence equating all the coefficients of ${x^3}$ term,${x^2}$ term and $x$ term to zero:
$ \Rightarrow A + C = 0$
\[ \Rightarrow \sqrt 2 A + B - \sqrt 2 C + D = 0\]
$ \Rightarrow A + \sqrt 2 B + C - \sqrt 2 D = 0$
$ \Rightarrow B + D = 1$
Now we have four equations and four variables, hence solving for the values of $A,B,C$ and $D$:
$ \Rightarrow A = \dfrac{{ - 1}}{{2\sqrt 2 }},B = \dfrac{1}{2},C = \dfrac{1}{{2\sqrt 2 }},D = \dfrac{1}{2}$
Now substituting the values of $A,B,C$ and $D$ in the partial fractions as shown:
$ \Rightarrow \dfrac{1}{{{x^4} + 1}} = \dfrac{{\dfrac{{ - 1}}{{2\sqrt 2 }}x + \dfrac{1}{2}}}{{\left( {{x^2} - \sqrt 2 x + 1} \right)}} + \dfrac{{\dfrac{1}{{2\sqrt 2 }}x + \dfrac{1}{2}}}{{\left( {{x^2} + \sqrt 2 x + 1} \right)}}$

\[ \Rightarrow \dfrac{1}{{{x^4} + 1}} = \dfrac{{ - \sqrt 2 x + 2}}{{4\left( {{x^2} - \sqrt 2 x + 1} \right)}} + \dfrac{{\sqrt 2 x + 2}}{{4\left( {{x^2} + \sqrt 2 x + 1} \right)}}\]

Note:
Please note that partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. Partial fractions are a way of breaking apart fractions with polynomials in them. The process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called partial fraction decomposition.